Question

Find the probability of a random sample with a normally distribution mean μ and variance σ^2 has value between μ - 0.54σ and μ + 1.72σ? What is the percentages off all possible outcomes of a normal random sample lie within plus or minus 0.64 SD's of the mean? Show work please.

Answer 1

P (μ - 0.54σ <= X <= μ + 1.72 σ) = P ( (μ - 0.54σ - μ ) / σ <= (X - μ ) / σ <= (μ + 1.72 σ - μ ) / σ ) = P (-0.54 <= Z <= 1.72)

= P(Z <= 1.72) - P(Z <= -0.54) = 0.957284 - 0.294599 = 0.662685

On similar lines, P (μ - 64σ <= X <= μ + 0.64σ) = P (-0.64 <= Z <= 0.64) = P(Z <= 0.64) - P(Z <= -0.64) = 0.738914 - 0.261086 = 0.477827

So we can conclude that about 47.78% of all possible outcomes lie within plus or minus 0.64 SD's of the mean